120       VI   HUBERT SPACES
such that \\x — y\\ = d(x, F). The point y = PF(x) is also the only point z e F
such that x - z is orthogonal to F. The mapping x -> PF(x) 0/ E onto F w
/we<2r, continuous, and of norm 1 z/ F ^ {0}; its kernel F' = P~1(Q) is the
subspace orthogonal to ¥,andE is the topological direct sum (see Section (5.4))
o/F and F'. Finally, F w rte subspace orthogonal to F'.
Let a = rf(je, F); by definition, there exists a sequence (yn) of points
of F such that   lim ||x — j>J = a; we prove (yn) is a Cauchy sequence.
h-+oo
Indeed, for any two points u, v of E, it follows from (6.1 .1) that
(6.3.1 .1)	\\u + v\\2 + \\u - v\\2 = 2(N|2 + ||i?||2)
hence \\ym - yj2 = 2(||* - ya\\2 + ||x - jj2) - 4||* - KyM + 7«)ll2. But
iC^m + 7J e F, hence \\x - %(ym + yn)\\2 ^ a2 ; therefore, if nQ is such that for
n ^ «0 , \\x - yn\\2 < a2 + e, we have, for m ^ w0 and n ^ «0 > Ibm - ^nii? < 4fi3
which proves our contention. As F is complete, the sequence (yn) tends
to a limit y e F, for which ||x - y\\ = d(x9 F). Suppose y' e F is also such
that \\x—y'\\=d(x,]?); using again (6.3.11), we obtain ||j>-/||2 =
4a2- 4 \\x - $(y + /)||2, and as %(y + /) e F, this implies ||j; - / 1|2 < 0, i.e.
yr - y. Let now z ^ 0 be any point of F, and write that \\x - ( y + Az)||2 > a2
for any real scalar A ^ 0; this, by (6.1 .1 ), gives
and this would yield a contradiction if we had $(x — y \ z) ^ 0, by a suitable
choice of L Hence 3%(x — y \ z) = 0, and replacing z by /z (if E is a complex
prehilbert space) shows that J(x — y \ z) = 0, hence (x — y \ z) = 0 in
every case; in other words x — y is orthogonal to F. Let y' e F be such
that x — / is orthogonal to F; then, for any z^O in F, we have
\\x - (/ + z)||2 = ||x - / 1|2 + ||z||2 by Pythagoras' theorem, and this proves
that / = y by the previous characterization of y.
This last characterization of y = PF(x) proves that PF is linear, for if
x — y and x' — y' are orthogonal to F, then Xx — ly is orthogonal to F and
so is (x + x') — (y + /) = (x - y) + (x' - /); as y + / e F and ky e F, this
shows that y + / = P¥(x + x') and ly = PF(Ax). By Pythagoras' theorem, we
have
(6.3.1.2)	\\x\\2 = \\PF(x)\\2 + ||^ - ?F(x)||2
and this proves that ||PF(x)|| < ||x||, hence (5.5.1) PF is continuous and has
norm ^ 1 ; but as PF(X) = x for x e F, we have ||PF|| = 1 if F is not reduced
to 0. The definition of PF implies that F' = Pf *(0) consists of the vectors x
orthogonal to F; as x = PF(;c) + (x — PF(x)) and x - PF(x) e Fr for any
x e E, we have E = F + F'; moreover, if x e F n F', x is isotropic, hence